How to delete files named like these? [closed] - linux

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I have two files in my home folder named "'?" and "'?;?" (without the double quotes). How can I delete them? I've tried to use escape, but it doesn't work.

Answer 1

Use single or double quotes to avoid wildcard expansion. A ? is a wildcard which indicates to the shell to match with any one single character. By placing it in quotes you are telling the shell not to perform wildcard expansion.
rm '?' '?;?'
rm "?" "?;?"
This will remove the two files named "?" and "?;?"
You can also use a backslash to quote the individual characters that have special meaning to the shell, so you could do this
rm \? \?\;\?
Notice you have to quote the '?' to prevent pathname expansion and you have to quote the ';' so the shell doesn't interpret that as separating commands.
If you leave out the quotes, then the shell parses it differently. Here's an experiment I ran.
$ for i in {1..4}; do for j in {a..c}; do touch "$i;$j" $j '?' '?;?';done;done
$ ls
1;a 1;b 1;c 2;a 2;b 2;c 3;a 3;b 3;c 4;a 4;b 4;c ? ?;? a b c
$ rm ? ?;?
rm: cannot remove `?': No such file or directory
rm: cannot remove `a': No such file or directory
rm: cannot remove `b': No such file or directory
rm: cannot remove `c': No such file or directory
bash: ?: command not found
$ rm `echo "?" "?;?"`
rm: cannot remove `?': No such file or directory
$
What happened here is the shell did pathname expansion, so
rm ? ?;?
became
rm ? a b c ? a b c;? a b c
The rm command removed files a b c ? then complained that the following files were not found (they had already been deleted). The semicolon separated commands, so it then tried to invoke the '?' command passing arguments "a" "b" "c" ... but there is no '?' command - the file named '?' had just been deleted, and it wasn't executable anyway - so the shell complains that the "?" command is not found.
If you want to remove all files matching "?" and "?;?" you need to trick the shell into expanding those, which I did like this
rm `echo "?" "?;?"`
This was expanded by the shell in two steps, first it runs echo "?" "?;?" which results in two strings, "?" and "?;?", then it does pathname expansion using those strings to produce the arguments for rm, which results in
rm ? 1;a 1;b 1;c 2;a 2;b 2;c 3;a 3;b 3;c 4;a 4;b 4;c ?;?
Notice that the wildcard expansion for '?' didn't produce any matching files this time (they had already been previously deleted), so the shell passes '?' as an argument to rm, which successfully removes all files passed as arguments except for '?' so it complains about that.
Here's another experiment
$ for i in {1..4}; do for j in {a..c}; do touch "$i;$j" $j '?' '?;?';done;done
$ ls
1;a 1;b 1;c 2;a 2;b 2;c 3;a 3;b 3;c 4;a 4;b 4;c ? ?;? a b c
$ rm "?" "?;?"
$ ls
1;a 1;b 1;c 2;a 2;b 2;c 3;a 3;b 3;c 4;a 4;b 4;c a b c
$ rm `echo "?" "?;?"`
$ ls
$
For more information consult the man page on globbing
man 7 glob
Wildcard Matching
A string is a wildcard pattern if it contains one of the characters '?', '*' or '['. Globbing is the operation that
expands a wildcard pattern into the list of pathnames matching the pattern. Matching is defined by:
A '?' (not between brackets) matches any single character.
A '*' (not between brackets) matches any string, including the empty string.

Answer 2

Note that ls can report a question mark for arbitrary non-printable characters, so there's a chance that what you've got as a file name does not contain a question mark.
You can spot this with the ls -b command, or with ls | cat.
As a convoluted example, complete with remedy, I created a script convolvulus like this:
set -x
mkdir convoluted &&
(
cd convoluted
cp /dev/null "$(ls -la | sed 1d)"
ls
ls -b
ls | cat
ls -la | cat
cp /dev/null $'\n'
cp /dev/null $'\n;\n'
ls -als | cat
ls -lab
ls
ls | cat
rm $'\n' $'\n;\n' d*
ls -a
)
rm -fr convoluted
When run, it yielded:
$ bash convolvulus 2>&1 | so
+ mkdir convoluted
+ cd convoluted
++ ls -la
++ sed 1d
+ cp /dev/null 'drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..'
+ ls
drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ ls -b
drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .\ndrwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ ls
+ cat
drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ ls -la
+ cat
total 0
drwxr-xr-x 3 jleffler staff 102 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
-rw-r--r-- 1 jleffler staff 0 Mar 9 11:58 drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ cp /dev/null '
'
+ cp /dev/null '
;
'
+ ls -als
+ cat
total 0
0 -rw-r--r-- 1 jleffler staff 0 Mar 9 11:58
0 -rw-r--r-- 1 jleffler staff 0 Mar 9 11:58
;
0 drwxr-xr-x 5 jleffler staff 170 Mar 9 11:58 .
0 drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
0 -rw-r--r-- 1 jleffler staff 0 Mar 9 11:58 drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ ls -lab
total 0
-rw-r--r-- 1 jleffler staff 0 Mar 9 11:58 \n
-rw-r--r-- 1 jleffler staff 0 Mar 9 11:58 \n;\n
drwxr-xr-x 5 jleffler staff 170 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
-rw-r--r-- 1 jleffler staff 0 Mar 9 11:58 drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .\ndrwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ ls
;
drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ ls
+ cat
;
drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..
+ rm '
' '
;
' 'drwxr-xr-x 2 jleffler staff 68 Mar 9 11:58 .
drwxr-xr-x 240 jleffler staff 8160 Mar 9 11:58 ..'
+ ls -a
.
..
+ rm -fr convoluted
$
Have fun!
The -b option to ls works for GNU ls and for Mac OS X and BSD ls (but is not defined by POSIX).
The '$'\n' notation is Bash ANSI-C Quoting.

Answer 3

Use quotes around the file names:
$ ls
? ?;?
$ rm '?'
$ ls
?;?
$ rm "?;?"
$ ls
$